Prove the following identities

1*2^0 + 2*2^1 + 3*2^2 + . . . + n*2^2n-1= (n-1)2^n + 1

Also

1 + 3 + 9 + 27 + . . . + 3^n-1=(3n-1)

2

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- Sep 13th 2010, 05:45 PMtaylor1234Proving Identities
Prove the following identities

1*2^0 + 2*2^1 + 3*2^2 + . . . + n*2^2n-1= (n-1)2^n + 1

Also

1 + 3 + 9 + 27 + . . . + 3^n-1=__(3n-1)__

2 - Sep 13th 2010, 05:51 PMProve It
I assume the second one should be

.

This is a geometric series with .

So

. - Sep 13th 2010, 08:05 PMSoroban
Hello, taylor1234!

There is a typo in the first one . . . the general term is not correct.

Quote:

In each term, the exponent is one less than the coefficient: .

We have: . . . . .

Multiply by 2: .

Subtract [2] - [1]: .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

- Sep 14th 2010, 07:16 AMtaylor1234
How would I solve these problems using the induction method. Assuming Sn is true and testing for Sn+1?